The Stacks project

Proof. Let $\xi \in Y$ be the generic point. Let $X_\xi $ be the scheme theoretic fibre of $\varphi $ over $\xi $. Denote $\mathcal{L}_\xi $ the pullback of $\mathcal{L}$ to $X_\xi $. Assumption (5) means that $\mathcal{L}_\xi $ is trivial. Choose a trivializing section $s \in \Gamma (X_\xi , \mathcal{L}_\xi )$. Observe that $X$ is integral by Lemma 31.11.7. Hence we can think of $s$ as a regular meromorphic section of $\mathcal{L}$. Pullbacks of meromorphic functions are defined for $\varphi $ by Lemma 31.23.5. Let $\mathcal{N} \subset \mathcal{K}_ Y$ be the $\mathcal{O}_ Y$-module whose sections over an open $V \subset Y$ are those meromorphic functions $g \in \mathcal{K}_ Y(V)$ such that $\varphi ^*(g)s \in \mathcal{L}(\varphi ^{-1}V)$. A priori $\varphi ^*(g)s$ is a section of $\mathcal{K}_ X(\mathcal{L})$ over $\varphi ^{-1}V$. We claim that $\mathcal{N}$ is an invertible $\mathcal{O}_ Y$-module and that the map

is an isomorphism.

We first prove the claim in the following situation: $X$ and $Y$ are affine and $\mathcal{L}$ trivial. Say $Y = \mathop{\mathrm{Spec}}(R)$, $X = \mathop{\mathrm{Spec}}(A)$ and $s$ given by the element $s \in A \otimes _ R K$ where $K$ is the fraction field of $R$. We can write $s = a/r$ for some nonzero $r \in R$ and $a \in A$. Since $s$ generates $\mathcal{L}$ on the generic fibre we see that there exists an $s' \in A \otimes _ R K$ such that $ss' = 1$. Thus we see that $s = r'/a'$ for some nonzero $r' \in R$ and $a' \in A$. Let $\mathfrak p_1, \ldots , \mathfrak p_ n \subset R$ be the minimal primes over $rr'$. Each $R_{\mathfrak p_ i}$ is a discrete valuation ring (Algebra, Lemmas 10.60.11 and 10.157.4). By assumption $\mathfrak q_ i = \mathfrak p_ i A$ is a prime. Hence $\mathfrak q_ i A_{\mathfrak q_ i}$ is generated by a single element and we find that $A_{\mathfrak q_ i}$ is a discrete valuation ring as well (Algebra, Lemma 10.119.7). Of course $R_{\mathfrak p_ i} \to A_{\mathfrak q_ i}$ has ramification index $1$. Let $e_ i, e'_ i \geq 0$ be the valuation of $a, a'$ in $A_{\mathfrak q_ i}$. Then $e_ i + e'_ i$ is the valuation of $rr'$ in $R_{\mathfrak p_ i}$. Note that

in $R$ by Algebra, Lemma 10.157.6. Set

so that $II' \subset (rr')$. Observe that

by Algebra, Lemmas 10.64.3 and 10.39.2. Similarly for $I'A$. Hence $a \in IA$ and $a' \in I'A$. We conclude that $IA \otimes _ A I'A \to rr'A$ is surjective. By faithful flatness of $R \to A$ we find that $I \otimes _ R I' \to (rr')$ is surjective as well. It follows that $II' = (rr')$ and $I$ and $I'$ are finite locally free of rank $1$, see Algebra, Lemma 10.120.16. Thus Zariski locally on $R$ we can write $I = (g)$ and $I' = (g')$ with $gg' = rr'$. Then $a = ug$ and $a' = u'g'$ for some $u, u' \in A$. We conclude that $u, u'$ are units. Thus Zariski locally on $R$ we have $s = ug/r$ and the claim follows in this case.

Let $y \in Y$ be a point. Pick $x \in X$ mapping to $y$. We may apply the result of the previous paragraph to $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$. We conclude there exists an element $g \in R(Y)^*$ well defined up to multiplication by an element of $\mathcal{O}_{Y, y}^*$ such that $\varphi ^*(g)s$ generates $\mathcal{L}_ x$. Hence $\varphi ^*(g)s$ generates $\mathcal{L}$ in a neighbourhood $U$ of $x$. Suppose $x'$ is a second point lying over $y$ and $g' \in R(Y)^*$ is such that $\varphi ^*(g')s$ generates $\mathcal{L}$ in an open neighbourhood $U'$ of $x'$. Then we can choose a point $x''$ in $U \cap U' \cap \varphi ^{-1}(\{ y\} )$ because the fibre is irreducible. By the uniqueness for the ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x''}$ we find that $g$ and $g'$ differ (multiplicatively) by an element in $\mathcal{O}_{Y, y}^*$. Hence we see that $\varphi ^*(g)s$ is a generator for $\mathcal{L}$ on an open neighbourhood of $\varphi ^{-1}(y)$. Let $Z \subset X$ be the set of points $z \in X$ such that $\varphi ^*(g)s$ does not generate $\mathcal{L}_ z$. The arguments above show that $Z$ is closed and that $Z = \varphi ^{-1}(T)$ for some subset $T \subset Y$ with $y \not\in T$. If we can show that $T$ is closed, then $g$ will be a generator for $\mathcal{N}$ as an $\mathcal{O}_ Y$-module in the open neighbourhood $Y \setminus T$ of $y$ thereby finishing the proof (some details omitted).

If $\varphi $ is quasi-compact, then $T$ is closed by Morphisms, Lemma 29.25.12. If $\varphi $ is locally of finite type, then $\varphi $ is open by Morphisms, Lemma 29.25.10. Then $Y \setminus T$ is open as the image of the open $X \setminus Z$. $\square$

Proof. Let $D' \subset X$ be the scheme theoretic image of the morphism $D \to X$. Since $X$ is locally Noetherian the morphism $D \to X$ is quasi-compact, see Properties, Lemma 28.5.3. Hence the formation of $D'$ commutes with passing to opens in $X$ by Morphisms, Lemma 29.6.3. Thus we may assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Let $I \subset A$ be the ideal corresponding to $D'$. Let $\mathfrak p \subset A$ be a prime ideal corresponding to a point of $X \setminus U$. To finish the proof it is enough to show that $I_\mathfrak p$ is generated by one element, see Lemma 31.15.2. Thus we may replace $X$ by $\mathop{\mathrm{Spec}}(A_\mathfrak p)$, see Morphisms, Lemma 29.25.16. In other words, we may assume that $X$ is the spectrum of a local UFD $A$. Then all local rings of $A$ are UFD's. It follows that $D = \sum a_ i D_ i$ with $D_ i \subset U$ an integral effective Cartier divisor, see Lemma 31.15.11. The generic points $\xi _ i$ of $D_ i$ correspond to prime ideals $\mathfrak p_ i \subset A$ of height $1$, see Lemma 31.15.3. Then $\mathfrak p_ i = (f_ i)$ for some prime element $f_ i \in A$ and we conclude that $D'$ is cut out by $\prod f_ i^{a_ i}$ as desired. $\square$

Proof. Choose $x \in X$, $x \not\in U$. We will show there exists an affine open neighbourhood $W \subset X$, such that $\mathcal{L}|_{W \cap U}$ extends to an invertible sheaf on $W$. This implies by glueing of sheaves (Sheaves, Section 6.33) that we can extend $\mathcal{L}$ to the strictly bigger open $U \cup W$. Let $W = \mathop{\mathrm{Spec}}(A)$ be an affine open neighbourhood. Since $U \cap W$ is quasi-affine, we see that we can write $\mathcal{L}|_{W \cap U}$ as $\mathcal{O}(D_1) \otimes \mathcal{O}(D_2)^{\otimes -1}$ for some effective Cartier divisors $D_1, D_2 \subset W \cap U$, see Lemma 31.15.12. Then $D_1$ and $D_2$ extend to effective Cartier divisors of $W$ by Lemma 31.28.2 which gives us the extension of the invertible sheaf.

If $X$ is Noetherian (which is the case most used in practice), the above combined with Noetherian induction finishes the proof. In the general case we argue as follows. First, because every local ring of a point outside of $U$ is a domain and $X$ is locally Noetherian, we see that the closure of $U$ in $X$ is open. Thus we may assume that $U \subset X$ is dense and schematically dense. Now we consider the set $T$ of triples $(U', \mathcal{L}', \alpha )$ where $U \subset U' \subset X$ is an open subscheme, $\mathcal{L}'$ is an invertible $\mathcal{O}_{U'}$-module, and $\alpha : \mathcal{L}'|_ U \to \mathcal{L}$ is an isomorphism. We endow $T$ with a partial ordering $\leq $ defined by the rule $(U', \mathcal{L}', \alpha ) \leq (U'', \mathcal{L}'', \alpha ')$ if and only if $U' \subset U''$ and there exists an isomorphism $\beta : \mathcal{L}''|_{U'} \to \mathcal{L}'$ compatible with $\alpha $ and $\alpha '$. Observe that $\beta $ is unique (if it exists) because $U \subset X$ is dense. The first part of the proof shows that for any element $t = (U', \mathcal{L}', \alpha )$ of $T$ with $U' \not= X$ there exists a $t' \in T$ with $t' > t$. Hence to finish the proof it suffices to show that Zorn's lemma applies. Thus consider a totally ordered subset $I \subset T$. If $i \in I$ corresponds to the triple $(U_ i, \mathcal{L}_ i, \alpha _ i)$, then we can construct an invertible module $\mathcal{L}'$ on $U' = \bigcup U_ i$ as follows. For $W \subset U'$ open and quasi-compact we see that $W \subset U_ i$ for some $i$ and we set

For the transition maps we use the $\beta $'s (which are unique and hence compose correctly). This defines an invertible $\mathcal{O}$-module $\mathcal{L}'$ on the basis of quasi-compact opens of $U'$ which is sufficient to define an invertible module (Sheaves, Section 6.30). We omit the details. $\square$

Proof. Since $R$ is a UFD so is any localization of it and any polynomial ring over it (Algebra, Lemma 10.120.10). Thus if $U \subset \mathbf{A}^ n_ R$ is open, then the map $\mathop{\mathrm{Pic}}\nolimits (\mathbf{A}^ n_ R) \to \mathop{\mathrm{Pic}}\nolimits (U)$ is surjective by Lemma 31.28.3. The vanishing of $\mathop{\mathrm{Pic}}\nolimits (\mathbf{A}^ n_ R)$ is equivalent to the vanishing of the picard group of the UFD $R[x_1, \ldots , x_ n]$ which is proved in More on Algebra, Lemma 15.117.3. $\square$

Proof. Observe that the local rings of $X = \mathbf{P}^ n_ R$ are UFDs because $X$ is covered by affine pieces isomorphic to $\mathbf{A}^ n_ R$ and $R[x_1, \ldots , x_ n]$ is a UFD (Algebra, Lemma 10.120.10). Hence $X$ is an integral Noetherian scheme all of whose local rings are UFDs and we see that $\mathop{\mathrm{Pic}}\nolimits (X) = \text{Cl}(X)$ by Lemma 31.27.7.

The displayed map is a group homomorphism by Constructions, Lemma 27.10.3. The map is injective because $H^0$ of $\mathcal{O}_ X$ and $\mathcal{O}_ X(m)$ are non-isomorphic $R$-modules if $m > 0$, see Cohomology of Schemes, Lemma 30.8.1. Let $\mathcal{L}$ be an invertible module on $X$. Consider the open $U = D_+(T_0) \cong \mathbf{A}^ n_ R$. The complement $H = X \setminus U$ is a prime divisor because it is isomorphic to $\text{Proj}(R[T_1, \ldots , T_ n])$ which is integral by the discussion in the previous paragraph. In fact $H$ is the zero scheme of the regular global section $T_0$ of $\mathcal{O}_ X(1)$ hence $\mathcal{O}_ X(1)$ maps to the class of $H$ in $\text{Cl}(X)$. By Lemma 31.28.4 we see that $\mathcal{L}|_ U \cong \mathcal{O}_ U$. Let $s \in \mathcal{L}(U)$ be a trivializing section. Then we can think of $s$ as a regular meromorphic section of $\mathcal{L}$ and we see that necessarily $\text{div}_\mathcal {L}(s) = m[H]$ for some $m \in \mathbf{Z}$ as $H$ is the only prime divisor of $X$ not meeting $U$. In other words, we see that $\mathcal{L}$ and $\mathcal{O}_ X(m)$ map to the same element of $\text{Cl}(X)$ and hence $\mathcal{L} \cong \mathcal{O}_ X(m)$ as desired. $\square$